Exactness and the Member Calculus — Examples

In $\mathbf{A}\mathbf{b}$, subobjects of $A$ are exactly subgroup inclusions into $A$, so [L2] identifies the class of $x$ with the subgroup represented by its image inclusion.

Therefore the abstract member-subobject correspondence is, in this category, the ordinary rule $x\mapsto \mathrm{im}(x)$.

The map $x_a$ is a group homomorphism $\mathbb{Z}\to A$, hence a member of $A$ by [L1].

Its image is the subgroup $⟨a⟩$, so [L2] identifies the corresponding subobject with the cyclic subgroup generated by $a$.

The image of the member $1_{{\mathbb{Z}}^2}$ is all of ${\mathbb{Z}}^2$. By [L1], its equivalence class corresponds to the whole subgroup ${\mathbb{Z}}^2\le {\mathbb{Z}}^2$.

Any member coming from a map $\mathbb{Z}\to {\mathbb{Z}}^2$ has cyclic image, because the image of $\mathbb{Z}$ is generated by the image of $1$. The subgroup ${\mathbb{Z}}^2$ is not cyclic. Therefore $1_{{\mathbb{Z}}^2}$ is not equivalent to any member $\mathbb{Z}\to {\mathbb{Z}}^2$.

This refutes the statement.

If $ix\equiv 0$, choose an epic cover $u:W\to X$ with $ixu=0$. Since $i$ is the subgroup inclusion, $xu=0$. Hence $x\equiv 0$.

By [L1], this proves that $i$ is monic; by [L2], it is the same computation as cancellation of equal members after applying $i$.

If $qh=0$, then every value of $h$ is even, so $h(W)\subseteq 2\mathbb{Z}$. Therefore there is a unique homomorphism $l:W\to 2\mathbb{Z}$ with $h=il$.

This realizes the cover in [L1] with $V=W$ and $k=1_W$. So the exactness criterion is visible here without any nontrivial refinement of domains.

Here $\ker (f)=0$, $\ker (gf)=\mathbb{Z}$, $\ker (g)=2\mathbb{Z}$, $\mathrm{coker}(f)\cong \mathbb{Z}\mathrm{/}2$, $\mathrm{coker}(gf)\cong \mathbb{Z}\mathrm{/}2$, and $\mathrm{coker}(g)=0$.

The map $\ker (gf)\to \ker (g)$ is just $f$, so it is multiplication by $2$ onto the subgroup $2\mathbb{Z}$. The connecting map $\ker (g)\to \mathrm{coker}(f)$ is zero because it is induced by the cokernel map $q_f:\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$, which kills every even integer. The map $\mathrm{coker}(f)\to \mathrm{coker}(gf)$ is the identity on $\mathbb{Z}\mathrm{/}2$, and the remaining arrows are the obvious zero maps. This is the displayed sequence.

That concrete sequence is exact by direct inspection.

The sequence is short exact in $\mathbf{A}\mathbf{b}$ by the usual kernel-image computation.

If $q$ had a section $s$, then $x:=s(1+2\mathbb{Z})$ would be an odd integer with $2x=s(0)=0$, impossible in $\mathbb{Z}$. So no section exists.

By [L2], the short exact sequence is nonsplit. This refutes the statement.

A short exact sequence of modules is a short exact sequence in an abelian category, and the given section or retraction is exactly the additional datum required by [L1].

The complementary map produced by [L1] satisfies $ps=1$, $ri=1$, and $ir+sp=1$, exactly the module equations stated in [L2].

Therefore the module theorem is the concrete $R\text{-}\mathbf{M}\mathbf{o}\mathbf{d}$ instance of the categorical one.

The vertical kernels are $0$, $0$, and $\mathbb{Z}\mathrm{/}2$, so the kernel row is exactly $0\to 0\to 0\to \mathbb{Z}\mathrm{/}2$.

The last arrow is the zero map $0\to \mathbb{Z}\mathrm{/}2$, hence not epic.

So the row cannot be short exact.